Complex Matrix Limit If $A$ is an $n \times n$ complex matrix, show that if $\lim_{k\rightarrow\infty}||A^kv||=0$ for every vector $v \in \Bbb C^n$, then $|\lambda|\leq1$ for every eigenvalue $\lambda$ of $A$.
 A: Proof by contrapositive:  suppose there exists an eigenvalue $\lambda$ with $|\lambda| > 1$, and corresponding eigenvector $\mathbf{v}$.  Consider $\lim_{k\to\infty} \|A^k \mathbf{v}\|$.
A: Suppose $\Vert A^k v \Vert \to 0$ as $k \to \infty$ for every $v \in \Bbb C^n$.  If $0 \ne w \in C^n$ is an eigenvector corresponding to eigenvalue $\lambda$, $A w = \lambda w$, we have
$A^m w = \lambda^m w,\tag{1}$
for every positive integer $m$, since
$A^2 w = A(Aw) = A(\lambda w) = \lambda (Aw) = \lambda^2 w, \tag{2}$
$A^3 w = A(A^2w) = A(\lambda^2 w) = \lambda^2 (Aw) = \lambda^3 w, \tag{3}$
and so forth.  Normalizing $w$, so that $\Vert w \Vert = 1$, we see from (1) that
$\Vert A^k w \Vert = \Vert \lambda^k w \Vert = \vert \lambda \vert^k \Vert w \Vert = \vert \lambda \vert^k, \tag{4}$
and
$\vert \lambda \vert^k \to 0 \; \text{as} \; k \to \infty \tag{5}$
if and only if
$\vert \lambda \vert < 1. \tag{6}$
Note we have actually shown something stronger than requested, i.e. that $\vert \lambda \vert < 1$, strictly.  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
